Is the diagonal of finitely presented groups computable? Let $f:G\to H$ be a surjective homomorphism of finitely presented groups. If the kernel of $f$ is finitely generated then is $G\times_H G$ is a finitely presented group? Can one compute an explicit finite presentation?
 A: The answer is "no".  The hypotheses for the fibre product to be finitely presentable are given by the 1-2-3 theorem of Bridson--Baumslag--Miller--Short:

1-2-3 Theorem (BBMS): The fibre product $G\times_H G$ is guaranteed to be finitely presentable as long as:

*

*the kernel of $G\to H$ is finitely generated;

*$G$ itself is finitely presented;

*$H$ satisfies a higher finiteness condition, namely being of "type $F_3$".


However, there is no general algorithm that can compute a finite presentation for $G\times_H G$. Indeed, stronger, there is no algorithm that can compute its abelianisation, and the result follows since the abelianisation can be computed from a finite presentation.
This is proved in my paper with Bridson:

Bridson & Wilton, "On the difficulty of presenting finitely presentable groups", Groups Geom. Dyn. 5(2), pp. 301–325, 2011

available on the arXiv here. See Theorem A, and note in the proof that the subgroup $\Lambda_n$ is indeed a fibre product.
To give a brief sketch of the argument, one takes a sequence of perfect groups $Q_n$ such that the second Betti numbers $b_2(Q_n)$ are known to be impossible to compute. The Rips short exact sequence gives
$$1\to K_n\to\Gamma_n\to Q_n\to 1$$
with $K_n$ finitely generated, and then let $\Lambda_n\leq \Gamma_n\times\Gamma_n$ be the fibre product. Finally, an argument with the Stallings exact sequence relates the Betti numbers:
$$b_1(\Lambda_n) = 2b_1(\Gamma_n)+b_2(Q_n).$$
Since $b_1(\Gamma_n)$ is computabe and $b_2(Q_n)$ isn't, it follows that $b_1(\Lambda_n)$ is not computable. In particular, no presentation for $\Lambda_n$ can be computed, even though we are given explicit generating sets and know that they are finitely presented.
